Plane Equation Derivation: Vector and Cartesian Forms

This solution details how to find both the vector and Cartesian equations of a plane. We are given that the plane passes through a specific point and is perpendicular to a line with known direction ratios. Understanding these concepts is crucial for mastering 3D geometry.

Point and Normal Vector Identification for Plane Equation

To find the equation of a plane, we need two key pieces of information:

1. A point through which the plane passes.
2. A normal vector to the plane.

From the question, we have:

• The plane passes through the point \((x_1, y_1, z_1) = (5, 2, -4)\).
• The plane is perpendicular to a line with direction ratios \(2, 3, -1\).

When a plane is perpendicular to a line, the direction ratios of that line serve as the components of the normal vector to the plane. Therefore, the normal vector, denoted as \(\vec{n}\), is:

\(\vec{n} = 2\hat{i} + 3\hat{j} - 1\hat{k} = 2\hat{i} + 3\hat{j} - \hat{k}\)

The position vector of the given point \((5, 2, -4)\) is \(\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}\).

Vector Equation of the Plane

The general vector equation of a plane passing through a point with position vector \(\vec{a}\) and having a normal vector \(\vec{n}\) is given by:

\(\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}\)

Where \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\) is the position vector of any point on the plane.

First, let's calculate the dot product \(\vec{a} \cdot \vec{n}\):

\[ \vec{a} \cdot \vec{n} = (5\hat{i} + 2\hat{j} - 4\hat{k}) \cdot (2\hat{i} + 3\hat{j} - \hat{k}) \] \[ = (5)(2) + (2)(3) + (-4)(-1) \] \[ = 10 + 6 + 4 \] \[ = 20 \]

Now, substitute the values into the vector equation:

\[ (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 20 \]

Thus, the vector equation of the plane is \(\vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 20\).

Cartesian Equation of the Plane

The Cartesian equation of a plane passing through a point \((x_1, y_1, z_1)\) and having direction ratios of the normal as \((a, b, c)\) is given by:

\[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \]

From our problem, we have:

• Point \((x_1, y_1, z_1) = (5, 2, -4)\)
• Normal vector components \((a, b, c) = (2, 3, -1)\)

Substitute these values into the Cartesian equation formula:

\[ 2(x - 5) + 3(y - 2) + (-1)(z - (-4)) = 0 \]

Simplify the equation step-by-step:

\[ 2(x - 5) + 3(y - 2) - 1(z + 4) = 0 \]

Distribute the constants:

\[ 2x - 10 + 3y - 6 - z - 4 = 0 \]

Combine the constant terms:

\[ 2x + 3y - z - 10 - 6 - 4 = 0 \]

\[ 2x + 3y - z - 20 = 0 \]

Move the constant to the right side of the equation:

\[ 2x + 3y - z = 20 \]

This is the Cartesian equation of the plane.

Comparison with Options

Let's compare our derived Cartesian equation \(2x + 3y - z = 20\) with the given options:

Our derived Cartesian equation matches Option 2.